The probability distribution is the major part in the probability theory and the statistics. The probability distribution is used to determine the number of possibility for the occurrence of an event. The most commonly used probability distributions are the binomial distribution, geometric distribution, normal distribution and the gamma distribution. These above mentioned distributions are included in the discrete and continuous probability distribution. The major type of the probability distribution is the discrete probability distribution and the continuous probability distribution. This article has the study about the response distribution.

Examples for response distribution:

Example 1 to response distribution:

If X is normally distributed the mean value is 4 and its standard deviation is 3. Determine the value of P (0 = X = 7).

Solution:

The given ‘mu’ value is 4 and the standard deviation is 3.

Z = ‘(X- mu)/ sigma’

When X = 0, Z = ‘(0- 4)/ 3’

= -‘4/3’

= -1.33

When X = 7, Z = ‘(7- 4)/ 3’

= ‘3/3’

= 1

Therefore,

P (0 = X = 7) = P (-1.33 less than Z less than 1)

P (0 = X = 7) = P (0 less than Z less than 1.33) + P (0 less than Z less than 1) (due to symmetry property)

P (0 = X = 7) = (0. 9082- 0.5) + (0.8413 – 0.5)

P (0 = X = 7) = 0.4082 + 0.3413

P (0 = X = 7) = 0.7495

The value for P (0 = X = 7) is 0.7495.

Example 2 to response distribution:

The probability for destroying the target in only one time is 0.47. Compute the probability that it would be destroyed on the third attempt itself.

Solution:

The probability of destroying the target in one trial is p = 0.47.

The value of the q is calculated by q = 1-p

q = 1- 0.47

q = 0.53

By the geometric distribution, the probability for the success is calculated by using the formula

P(X =x) = q x p, the value of x is 0, 1, 2. . .

The target is destroyed at the third attempt, so x = 3.

P(X = 3) = (0. 53) 3 (0.47)

P(X = 3) = (0.1489) (0.47)

P(X = 3) = 0.0699

The probability for destroying the target at the third trial is 0.0699.

More example problems for response distribution:

A car hire firm has three cars. The number of demands for a car as a Poisson distribution with mean of 7.3. Find the proportion of days on which neither car is used and the proportion of days on which some demand is refused for the car.

Solution:

Let X denotes the number of demands for a car.

The given mean value is 7.3.

By the Poisson distribution

P[X = x] = ‘(e^ – lambda lambda^x)/ (x!)’

Proportion of the days on which neither car is used = P[X = 0] = ‘e^-7.3’ = 0.00067

Proportion of days on which some demand is refused = P[X greater than 3]

P[X greater than 3] = 1- P[X = 3]

P[X greater than 3] = 1- [P (0) +P (1) + P (2) +P (3)]

P[X greater than 3] = 1- ‘e^-7.3’ (1+ 7.3+ 26.645+ 64.84)

P[X greater than 3] = 1 – (0.00067) (99.785)

P[X greater than 3] = 1 – 0.06686

P[X greater than 3] = 0.93314

The proportion of days on which neither the car used is 0.00067. The proportion of days on which some demand refused is 0.93314.